Determine 3 Elements of Group H with Primes p & q

  • Thread starter Thread starter mansi
  • Start date Start date
  • Tags Tags
    Subgroup
AI Thread Summary
The discussion focuses on identifying three elements of a group H, formed by distinct primes p and q, under addition. The key insight is that if H contains both p and q, it must also include all integers due to their coprimality, leading to the conclusion that H would be equivalent to the group of integers Z. Various combinations of elements from the set {p, p+q, pq, p^q, q^p} are considered, with the argument that including p and q necessitates the inclusion of all integers. The example of p=2 and q=3 illustrates this point, showing that if both primes are in the group, then 1 must also be included, leading to the entire set of integers. Ultimately, the conclusion is that option e (p, pq, p^q) is the correct answer, as it aligns with the properties of the group.
mansi
Messages
61
Reaction score
0
let p and q be distinct primes. suppose that H is a proper subset of integers and H is a group under addition that contains exactly 3 elements of the set
{ p,p+q,pq, p^q , q^p}.
Determine which of the foll are the 3 elements in H
a. pq, p^q, q^p

b. P+q, pq,p^q

c. p, p+q, pq

d. p, p^q, q^p

e. p, pq, p^q
 
Physics news on Phys.org
Hint Euclid's algorithm: p^r and q^s are coprime so if H contains these two elements, then it contains 1, and hence is Z. Use this idea in several variations. Of course you could consider the group pZ
 
thanks sir, but could you please elaborate further.
i don't seem to get the idea...
 
If a group contains p, it contains np for all n in Z. So clearly e. forms the answer.

A group for instance cannot contain p and q if they are coprime and not be all of Z since there are a and b in Z such that ap+bq=1, hence the group contains all elements of Z.

And I tihnk you ought to ponder that for a while, cos I really have given you more information than I want to.
 
How about thinking about an example if you cant' see it:

p=2 q=3

If 2 and 3 are in the group, then so is -2 (inverses) and hence, so is 3-2=1 (composition)

If 1 is in there so is 1+1+1+..+1= n (composition) and n was arbitrary, also -n is in there (inverses again)
 
well...thanks a lot, sir! i figured it out...
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top